Problem: $f(n)=2\cdot(-3)^{{\,n}}$ Complete the recursive formula of $f(n)$. $f(1)=$
$f( 1)=2\cdot(-3)^{ 1}={-6}$ $f( 2)=2(-3)^{ 2}={18}$ $\dfrac{f( 2)}{f( 1)}=\dfrac{{18}}{{-6}}={-3}$ So the first term of the sequence is ${-6}$ and the common difference is ${-3}$. This is the recursive formula of the sequence: $\begin{cases} f(1)={-6} \\\\ f(n)=f(n-1)\cdot({-3}) \end{cases}$